Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
vector<int> mapa[100100];
vector<int> g[100100];
int h[100100], p[100100][17];
int lef[4 * 17 * 100100], rig[4 * 17 * 100100];
int root[100100], cont = 1, st[4 * 17 * 100100];
void build(int no, int l, int r) {
  if (l == r) return;
  lef[no] = cont++;
  rig[no] = cont++;
  build(lef[no], l, (l + r) / 2);
  build(rig[no], (l + r) / 2 + 1, r);
}
int update(int no, int l, int r, int pos) {
  int novo = cont++;
  st[novo] = st[no];
  if (l == r) {
    st[novo]++;
    return novo;
  }
  lef[novo] = lef[no];
  rig[novo] = rig[no];
  if (pos <= (l + r) / 2)
    lef[novo] = update(lef[novo], l, (l + r) / 2, pos);
  else
    rig[novo] = update(rig[novo], (l + r) / 2 + 1, r, pos);
  st[novo] = st[lef[novo]] + st[rig[novo]];
  return novo;
}
void dfs(int x, int pai) {
  p[x][0] = pai;
  for (int i = 1; i < 17; ++i) p[x][i] = p[p[x][i - 1]][i - 1];
  root[x] = root[pai];
  for (int i = 0; i < mapa[x].size(); ++i)
    root[x] = update(root[x], 1, m, mapa[x][i]);
  for (int i = 0; i < g[x].size(); ++i) {
    int y = g[x][i];
    if (y == pai) continue;
    h[y] = h[x] + 1;
    dfs(y, x);
  }
}
int lca(int x, int y) {
  if (h[x] < h[y]) swap(x, y);
  for (int i = 17 - 1; i >= 0; --i)
    if (h[p[x][i]] >= h[y]) x = p[x][i];
  if (x == y) return x;
  for (int i = 17 - 1; i >= 0; --i)
    if (p[x][i] != p[y][i]) x = p[x][i], y = p[y][i];
  return p[x][0];
}
void query(int x, int y, int tam) {
  int lc = lca(x, y);
  int k, k2;
  int a, b, l, r;
  vector<int> ans;
  for (int i = 1; i <= tam; ++i) {
    a = root[x], b = root[y], k = root[lc], k2 = root[p[lc][0]];
    if (st[a] + st[b] - st[k] - st[k2] < i) break;
    int aux = i, valor;
    l = 1, r = m;
    while (l < r) {
      valor = st[lef[a]] + st[lef[b]] - st[lef[k]] - st[lef[k2]];
      if (aux <= valor) {
        a = lef[a];
        b = lef[b];
        k = lef[k];
        k2 = lef[k2];
        r = (l + r) / 2;
      } else {
        aux -= valor;
        a = rig[a];
        b = rig[b];
        k = rig[k];
        k2 = rig[k2];
        l = (l + r) / 2 + 1;
      }
    }
    ans.push_back(l);
  }
  printf("%d", (int)ans.size());
  for (int i = 0; i < ans.size(); ++i) {
    printf(" %d", ans[i]);
  }
  printf("\n");
}
int main() {
  int x, y, tam;
  scanf("%d %d %d", &n, &m, &q);
  g[0].push_back(1);
  g[1].push_back(0);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &x);
    mapa[x].push_back(i);
  }
  build(0, 1, m);
  dfs(0, 0);
  for (int i = 0; i < q; ++i) {
    scanf("%d %d %d", &x, &y, &tam);
    query(x, y, tam);
  }
  return 0;
}